Finding Equidistant Coordinates

Let the coordinates of point C be \((x, y)\).

The condition is that point C is equidistant from points A(5, -6) and B(7, 8). This means the distance AC is equal to the distance BC.

Mathematically, we write this as \(AC = BC\). Squaring both sides, we get \(AC^2 = BC^2\).

Applying the Distance Formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\). Therefore, \(d^2 = (x_2-x_1)^2 + (y_2-y_1)^2\).

Using this, we set up the equation for \(AC^2\) and \(BC^2\):

\(AC^2 = (x-5)^2 + (y - (-6))^2 = (x-5)^2 + (y+6)^2\)

\(BC^2 = (x-7)^2 + (y-8)^2\)

Solving the Equation

Now, we set \(AC^2 = BC^2\):

\((x-5)^2 + (y+6)^2 = (x-7)^2 + (y-8)^2\)

Expand the terms:

\((x^2 - 10x + 25) + (y^2 + 12y + 36) = (x^2 - 14x + 49) + (y^2 - 16y + 64)\)

Simplify the equation by canceling \(x^2\) and \(y^2\) terms from both sides and combining constants:

\(-10x + 12y + 61 = -14x - 16y + 113\)

Rearrange the terms to group x and y:

\(-10x + 14x + 12y + 16y = 113 - 61\)

\(4x + 28y = 52\)

Divide the entire equation by 4 to simplify:

\(x + 7y = 13\)

This equation represents the locus of all points equidistant from A and B (the perpendicular bisector).

Testing the Options

We now test the given options to see which one satisfies the equation \(x + 7y = 13\).

• Option 1: (6, 3) => \(6 + 7(3) = 6 + 21 = 27 \neq 13\)
• Option 2: (6, 7) => \(6 + 7(7) = 6 + 49 = 55 \neq 13\)
• Option 3: (6, 1) => \(6 + 7(1) = 6 + 7 = 13\). This satisfies the equation.
• Option 4: (4, 3) => \(4 + 7(3) = 4 + 21 = 25 \neq 13\)

Therefore, the coordinates of point C must be (6, 1).